Method of DifferentiationmediumFree

Method of Differentiation: Question

JEE Maths question with a full step-by-step solution.

Question
If y=(sin⁡−1x)2+(cos⁡−1x)2y = (\sin^{-1}x)^2 + (\cos^{-1}x)^2, then (1−x2)d2ydx2−xdydx(1 - x^2)\dfrac{d^2 y}{dx^2} - x\dfrac{dy}{dx} is:
A−4-4
B44correct
C22
D00
Solution
Step 1: First derivative
y1=2sin⁡−1x1−x2−2cos⁡−1x1−x2  ⟹  1−x2 y1=2(sin⁡−1x−cos⁡−1x)y_1 = \frac{2\sin^{-1}x}{\sqrt{1 - x^2}} - \frac{2\cos^{-1}x}{\sqrt{1 - x^2}} \implies \sqrt{1 - x^2}\, y_1 = 2(\sin^{-1}x - \cos^{-1}x)
 Step 2: Differentiate again
1−x2 y2−x1−x2 y1=2(11−x2+11−x2)=41−x2\sqrt{1 - x^2}\, y_2 - \frac{x}{\sqrt{1 - x^2}}\, y_1 = 2\left(\frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - x^2}}\right) = \frac{4}{\sqrt{1 - x^2}}
 Step 3: Multiply through by 1−x2\sqrt{1 - x^2}
(1−x2)y2−xy1=4(1 - x^2)y_2 - x y_1 = 4
 Answer: (2)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Method of Differentiation · easy
If f′(x)=3x2+6f'(x) = \sqrt{3x^2+6} and y=f(x3)y = f(x^3), then at x=1x = 1, (dydx)\left(\dfrac{dy}{dx}\right) is equal to.
Method of Differentiation · easy
If yy is a function of xx and log⁡(x+y)=2xy\log(x+y) = 2xy, then y′(0)y'(0) is.
Method of Differentiation · easy
ddx[sin⁡x+cos⁡x1+sin⁡2x]\dfrac{d}{dx}\left[\dfrac{\sin x + \cos x}{\sqrt{1 + \sin 2x}}\right], where 0<x<π40 < x < \dfrac{\pi}{4}, is:
Method of Differentiation · easy
If y=1+x2+x41+x+x2y=\dfrac{1+x^{2}+x^{4}}{1+x+x^{2}} and dydx=ax+b\dfrac{dy}{dx}=ax+b, then the values of aa and bb are

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath